替换给定字符串和换行符之间的字符串

Question

我有一个文件，其中仅包含一次 search_for_me=12.21/13.31/14，后跟一个 line break。我想用 21.12/44.22/44 替换 12.21/13.31/14。如何实现？

<?php
/*
$string = 'April 15, 2003';
$pattern = '/(\w+) (\d+), (\d+)/i';
$replacement = '1,';
echo preg_replace($pattern, $replacement, $string)."\n\n\n";
//April1,2003
*/
$string = file_get_contents('test.conf');

$string= <<<EOT
bla bla bla

search_for_me=12.21/13.31/14

bla bla bla
EOT;

echo $string."\n\n";

$replace = "21.12/44.22/44";

$search = "/[^search_for_me=](.*)[^\n]/";
echo preg_replace($search,$replace,$string)."\n\n";

echo 'done';

Answer 1

这里我使用正则表达式进行搜索和替换，

正则表达式： /(search_for_me).*?\n/，这将匹配 search_for_me 直到 \n

替换: '=21.12/44.22/44'."\n" 这里 </code> 将包含第一个捕获组 <code>search_for_me.

Try this code snippet here

<?php
/*
$string = 'April 15, 2003';
$pattern = '/(\w+) (\d+), (\d+)/i';
$replacement = '1,';
echo preg_replace($pattern, $replacement, $string)."\n\n\n";
//April1,2003
*/
$string = file_get_contents('test.conf');

$string= <<<EOT
bla bla bla

search_for_me=12.21/13.31/14

bla bla bla
EOT;

echo $string."\n\n";

$replace = "21.12/44.22/44";

$search = "/(search_for_me)=.*?\n/";
echo preg_replace($search,'=21.12/44.22/44'."\n",$string)."\n\n";

Answer 2

试试这个：

$string= <<<EOT
bla bla bla

search_for_me=12.21/13.31/14

bla bla bla
EOT;

printf("-String without replace:\n\n%s\n\n", $string);

$replace = '21.12/44.22/44';

$pattern = '/(?<=search_for_me\=)(.*)/';

$new_string = preg_replace($pattern, $replace, $string);

printf("-String with replace:\n\n%s", $new_string);

我使用正后视HERE

替换给定字符串和换行符之间的字符串

Replace string between given string and line break

php

regex

pcre